I didn't find any "XYZ-Wings" -- I've read about those, but I don't have much practice at finding them. I did notice two (prosaic) "XY-Wings." And the last tough step (for me) involved two binary chains on the digit "4", which solved seven or eight cells all at once.

Now, there are two XYZ-wings here. An XYZ wing is a triple that spans two elements. One of the three possibilities must be present in all three squares. That possibility can then be excluded from squares that are neighbors of all of the three XYZ squares.

Squares R2C3, R2C2 and R7C3 are an XYZ-wing on <7>. R1C3 cannot have possibility <7>.

Squares R8C6, R8C4 and R4C6 are an XYZ-wing on <8>. R9C6 cannot have possibility <8>.

How does (simple) coloring work? For any value of a possibility, look in each element (row, column, block) of the puzzle to find where the possibility occurs twice. Draw a line connecting the two squares.

(Actually, I just use a separate piece of squared paper with a 9x9 empty grid to do this.)

When you are done, you may have a collection of "graphs", networks of lines that connect pairs of possibilities. Now, you can use two colors (say, red and green) to alternately label each square on the graph. Either the red squares have the possible value, or the green squares do.

In the grid above, I have labeled the squares that form one connected graph for the possibility <7> with R(ed) and G(reen).

Note that R2 has two Green squares, so the possibility <7> must be in the Red squares!

For the possibility <1>, there are three unconnected graphs, A, B, and C. I have labeled the alternate possibilities A,a, B,b, etc. Graph A has 5 squares, the other two have 3 squares each.

Now, take a look at R6 and C7. I concluded I could join graphs A and B. So, set B = A and b = a. Then, R6 and C7 have both possibilities: C cannot be <1>, c must be <1> (R5C4 = <1>), and the puzzle is solved!

There are two binary chains in the "4"s. Note the contradiction at r2c3: the "4+" is aligned with both an "=" and a "~". So we can eliminate "4" at r2c3, r9c2, r8c8, & r6c9 -- and we can place "4"s at r8c3 & r9c9. The rest is simple. dcb

PS Keith, I'm not sure (yet) that linking the two chains on "1" is logically valid. If you had an "A" opposite a "b" and an "a" opposite a "B" then I could see it. But in the grid you posted all I can find ia an "a" opposite a "B", meaning that "A or b" is always true. I don't see how to prove the other half, that "a or B" is always true.

I'll write more after I've had a chance to study your solution in more detail. Thanks for the XYZ Wing steps -- those are great!

I agree with David about your colouring solution using the 1's. Your A-a and B-b chains link in box 6 and what can be concluded is that both (a) and (B) can't be true, so either/both (A) and (b) are true but there is no cell that shares a group with cells with both those markers, so no exclusions can be made.

However, using the same candidate profile as contained in your last grid, there are three conjugate chains in 4's that can be linked in such a manner as to make an exclusion. They are labeled A-a, B-b and C-c.

If both A & C were true then both B & b would be false, so either/both of a & c must be true. That makes exclusions of 4 at r9c2 and r6c9. Now the A-a chain can be extended by one cell to include r3c9(A).

Now both ends of the B-b chain are linked to cells marked with (A), which we can conclude must be false and all the cells marked with (a) are true. Once the resulting assignments are made, there is an X-wing, in a different number, which solves the puzzle.